1996-01-25 - The cost of breaking RC4 with a 40 bit key.

Header Data

From: norm@netcom.com (Norman Hardy)
To: cypherpunks@toad.com
Message Hash: ab9ec3ac45ec34f2ad653aba747732fa0737fdd01c5cad6d95f56f4242a0e44c
Message ID: <ad2d658900021004e878@DialupEudora>
Reply To: N/A
UTC Datetime: 1996-01-25 19:14:47 UTC
Raw Date: Fri, 26 Jan 1996 03:14:47 +0800

Raw message

From: norm@netcom.com (Norman Hardy)
Date: Fri, 26 Jan 1996 03:14:47 +0800
To: cypherpunks@toad.com
Subject: The cost of breaking RC4 with a 40 bit key.
Message-ID: <ad2d658900021004e878@DialupEudora>
MIME-Version: 1.0
Content-Type: text/plain


I think that special hardware to break RC4 would require 256 bytes of
registers and only a few hundred control gates. Lets say 5000 transistors
per "module". You can put several hundred modules on a chip. Each module
can easily do one step in 5 ns. I havn't figured out what the attack would
be (known plain text etc.) and hardware to handle that might be more. In
mass production the marginal cost of such a chip might be $100.  Perhaps
trying one key requires 100 steps. I get the cost per key trial as follows:

(100  $/chip)(100 steps/trial)(5 (module*ns)/step)/
((10^9 ns/sec)(10^8 sec/(economic lifetime))(200 modules / chip))

10^(2+2+.7 - (9+8+2.3)) $/keytrial= 10^(-15+.4) $/keytrial
 = 2.5*10^(-15) $/keytrial

I compute the cost of breaking a 40 bit key as 2.5*10^(-3) $ or one quarter
of a cent.







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